acid-base reactions copyright © 1999 by harcourt brace & company all rights reserved. requests...

Acid-Base ReactionsAcid-Base Reactions

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

Strong Acid and Strong Base

HCl + NaOH

HCl + HHCl + H22O --->O --->HH33OO++ + Cl + Cl--

NaOH(aq) ---> NaOH(aq) ---> NaNa++(aq) + OH(aq) + OH--(aq)(aq)

HH33OO++ + Cl + Cl- - + + NaNa+++ + OHOH-- 2 H2 H22OO + + ClCl- - + + NaNa++

Net ionic Equation doesn’t show spectator ions!



H2O + H2O H3O+ + OH- K = 1 x 10-14

HH33OO++ + + OHOH-- 2 H2 H22OO

K = 1/ 1 x 10-14 = 1 x 1014

NEUTRALIZATION REACTION, pH = 7



Strong Acid and Weak Base

NHNH33(aq)(aq) + H + H22O(liq) O(liq) NHNH44++(aq)(aq) + + OHOH--(aq)(aq)

HCl + NH3

HH33OO++ + + OHOH-- 2 H 2 H22OO

HH33OO++ + + NHNH33 H H22O + NHO + NH44

++

Knet = Kb (1/Kw) = 1.8 x 109



STRONG Acid and Weak Base

Mixing equal molar quantities gives an acidic solution.

Weak Acid and STRONG Base

Mixing equal molar quantities gives a basic solution.



Weak Acid and Weak Base

pH depends on Kb and Ka of conjugate base and acid.

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Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Stomach Acidity &Stomach Acidity &Acid-Base ReactionsAcid-Base Reactions

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Example 18.1—Calculate pH

100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3

HCl + HCl + NHNH33 Cl Cl-- + NH + NH44

++

0.100 L x 0.10 mol/L = 0.0100 mol HCl =0.0100 mol Cl-

0.050 L x 0.20 mol/L = 0.0100 mol NH3

=0.0100 mol NH4+

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100 mL of 0.10 M HCl with 50 mL of 0.20 M NH3

HCl + HCl + NHNH33 Cl Cl-- + NH + NH44

++

0.0100 mol Cl- / 0.150 L solution = 0.0667 M Cl-

0.0100 mol NH4+

/ 0.150 L solution = 0.0667 M NH4+

HH22O + NHO + NH44++ HH33OO++ + + NHNH33

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Equilibria of NH4+ contributes to pH

EQUIL Ka = x (x) / (0.0667 – x)

Ka = 5.6 x 10-10 and x = 6.1 x 10-6 M

HH22O + NHO + NH44++ HH33OO++ + + NHNH33

Ka = [Ka = [HH33OO++ ] [ ] [NHNH33] / ] / [[NHNH44++] ]

pH = 5.21

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Acid-Base ReactionsAcid-Base ReactionsSection 18.4Section 18.4

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution? solution?

HBz + NaOH ---> NaHBz + NaOH ---> Na++ + Bz + Bz-- + H + H22OO

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the of benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final equivalence point. What is the pH of the final solution? solution?

HBz + NaOH ---> NaHBz + NaOH ---> Na++ + Bz + Bz-- + H + H22OO

CC66HH55COCO22H = HBzH = HBzBenzoate ion = Bz-

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Acid-Base ReactionsAcid-Base ReactionsSection 18.4Section 18.4

The product of the titration of benzoic acid, The product of the titration of benzoic acid, the benzoate ion, Bzthe benzoate ion, Bz--, is the conjugate base , is the conjugate base of a weak acid. of a weak acid. The final solution is basic.The final solution is basic.

BzBz- - + H+ H22O HBz + OHO HBz + OH--

The product of the titration of benzoic acid, The product of the titration of benzoic acid, the benzoate ion, Bzthe benzoate ion, Bz--, is the conjugate base , is the conjugate base of a weak acid. of a weak acid. The final solution is basic.The final solution is basic.

BzBz- - + H+ H22O HBz + OHO HBz + OH--

++

KKbb = 1.6 x 10 = 1.6 x 10-10-10

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Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions

Strategy —Strategy — find the conc. of the find the conc. of the conjugate base Bconjugate base B-- in the solution AFTER in the solution AFTER the titration, then calculate pH.the titration, then calculate pH.

This is a two-step problemThis is a two-step problem

1.1. stoichiometrystoichiometry of acid-base reaction of acid-base reaction

2.2. equilibrium calculationequilibrium calculation

Strategy —Strategy — find the conc. of the find the conc. of the conjugate base Bconjugate base B-- in the solution AFTER in the solution AFTER the titration, then calculate pH.the titration, then calculate pH.

This is a two-step problemThis is a two-step problem

1.1. stoichiometrystoichiometry of acid-base reaction of acid-base reaction

2.2. equilibrium calculationequilibrium calculation

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M solution of You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH to the equivalence point. benzoic acid with 0.100 M NaOH to the equivalence point. What is the pH of the final solution?What is the pH of the final solution?

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STOICHIOMETRY PORTIONSTOICHIOMETRY PORTION

1. Calc. moles of NaOH req’d1. Calc. moles of NaOH req’d

(0.100 L HBz)(0.025 M) = (0.100 L HBz)(0.025 M) = 0.0025 mol HBz0.0025 mol HBz

This requires This requires 0.0025 mol NaOH0.0025 mol NaOH

2.2. Calc. volume of NaOH req’dCalc. volume of NaOH req’d

0.0025 mol (1 L / 0.100 mol) = 0.0025 mol (1 L / 0.100 mol) = 0.025 L0.025 L

25 mL of NaOH req’d 25 mL of NaOH req’d


1. Calc. moles of NaOH req’d1. Calc. moles of NaOH req’d

(0.100 L HBz)(0.025 M) = (0.100 L HBz)(0.025 M) = 0.0025 mol HBz0.0025 mol HBz

This requires This requires 0.0025 mol NaOH0.0025 mol NaOH

2.2. Calc. volume of NaOH req’dCalc. volume of NaOH req’d

0.0025 mol (1 L / 0.100 mol) = 0.0025 mol (1 L / 0.100 mol) = 0.025 L0.025 L



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3. Moles of Bz3. Moles of Bz-- produced = moles HBz = produced = moles HBz = 0.0025 mol0.0025 mol

4. Calc. conc. of Bz4. Calc. conc. of Bz--

There are 0.0025 mol of BzThere are 0.0025 mol of Bz-- in a in a TOTAL SOLUTION TOTAL SOLUTION VOLUMEVOLUME of of 125 mL125 mL

[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M



3. Moles of Bz3. Moles of Bz-- produced = moles HBz = produced = moles HBz = 0.0025 mol0.0025 mol

4. Calc. conc. of Bz4. Calc. conc. of Bz--

There are 0.0025 mol of BzThere are 0.0025 mol of Bz-- in a in a TOTAL SOLUTION TOTAL SOLUTION VOLUMEVOLUME of of 125 mL125 mL

[Bz[Bz--] = 0.0025 mol / 0.125 L = ] = 0.0025 mol / 0.125 L = 0.020 M0.020 M


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EQUILIBRIUM PORTIONEQUILIBRIUM PORTION

BzBz- - + H+ H22O O HBz + OHHBz + OH- - KKbb = 1.6 x 10 = 1.6 x 10-10-10

[Bz[Bz--]] [HBz][HBz] [OH[OH--]]

initialinitial changechange

equilibequilib




initialinitial changechange

equilibequilib


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initialinitial 0.0200.020 00 00

changechange -x-x +x+x +x+x

equilibequilib 0.020 - x0.020 - x xx xx




initialinitial 0.0200.020 00 00

changechange -x-x +x+x +x+x



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Solving in the usual way, we findSolving in the usual way, we find

x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25





Solving in the usual way, we findSolving in the usual way, we find

x = [OHx = [OH--] = 1.8 x 10] = 1.8 x 10-6-6, pOH = 5.75, and , pOH = 5.75, and pH = 8.25pH = 8.25


Kb 1.6 x 10-10 = x2

0.020 - xKb 1.6 x 10-10 =

x2

0.020 - x

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HBzHBz + H+ H22O O HH33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so

[H[H33OO++] = 1 (K] = 1 (Kaa ) = 6.3 x 10 ) = 6.3 x 10-5-5

pH = 4.20pH = 4.20

HBzHBz + H+ H22O O HH33OO++ + Bz + Bz- - KKaa = 6.3 x 10 = 6.3 x 10-5-5

At the half-way point, [HBz] = [BzAt the half-way point, [HBz] = [Bz--], so], so

[H[H33OO++] = 1 (K] = 1 (Kaa ) = 6.3 x 10 ) = 6.3 x 10-5-5

pH = 4.20pH = 4.20

QUESTION:QUESTION: You titrate 100. mL of a 0.025 M You titrate 100. mL of a 0.025 M solution of benzoic acid with 0.100 M NaOH solution of benzoic acid with 0.100 M NaOH What is the pH at the half-way point?What is the pH at the half-way point?

][

]][[ 3

HA

AOHKa

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Henderson-Hasselbalch Henderson-Hasselbalch EquationEquation

Take the negative log of both sides of this Take the negative log of both sides of this equationequation

[H3O] [Acid]

[Conj. base]•Ka[H3O]

[Acid][Conj. base]

•Ka][

]][[ 3

HA

AOHKa

pH pKa - log[Acid]

[Conj. base]

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oror

This is called the Henderson-Hasselbalch equation.This is called the Henderson-Hasselbalch equation.

pH pKa + log[Conj. base]

[Acid]pH pKa + log

[Conj. base][Acid]

pH pKa - log[Acid]

[Conj. base]

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This shows that the pH is determined largely This shows that the pH is determined largely by the pKby the pKaa of the acid and then adjusted by of the acid and then adjusted by the ratio of acid and conjugate base.the ratio of acid and conjugate base.

pH pKa + log[Conj. base]

[Acid]pH pKa + log

[Conj. base][Acid]

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